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3 years ago

Michael puts 3 beads next to each other

1 answer:

mihalych1998 [28]3 years ago

6 0

When we multiply a whole number with a fraction, we can think of the whole numbers as "over one"...

5 3 5 15
3 * --- = --- * ---- = ----
6 1 6 6

We can further simplify the fraction, by dividing numerator and denominator by 3...

15 15 / 3 5
--- = ------- = ----
6 6 / 3 2

Since this is an improper fraction, we need to turn it into a mixed number...

5 2 + 2 + 1 2 2 1 1 1
---- = -------------- = --- + ---- + --- = 1 + 1 + ---- = 2 --- or 2.5
2 2 2 2 2 2 2

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A deer can run for 2 hours at the speed of 45 mph how far can It run in that time

Virty [35]

Answer:

90 mph

Step-by-step explanation:

if it can go 45 mph, in 2 hrs, it should go 90

45*2=90

7 0

2 years ago

Read 2 more answers

) You are finally well!  But now your child is sick!  You know there have been mistakes in orders so you are on the internet t

rjkz [21]

Answer:

The doctor's dosage was not appropriate.

The right dosage is from 56.625 mg to 113.25 mg of Antibiotic to be given every 6 hour for the child with weight 25 lbs.

Step-by-step explanation:

It is given that 20 to 40 mg/kg/day is the recommended dosage.

Now, the doctor's order is 150 mg of antibiotic to be given every 6 hours. And my child weighs 25 lbs.

Now, 1 lb is equivalent to 0.453 kg.

So, my child's weight is (25 × 0.453) = 11.325 kg.

So, the doctor's order is 150 mg of antibiotic to be given every 6 hours for a child of weight 11.325 kg.

Hence, the dosage is [(150 × 4) ÷ 11.325] = 52.98 mg/kg/day.

So, this is not within the limit of 20 to 40 mg/kg/day.

Therefore, the doctor's dosage was not appropriate.

Now, let the appropriate dosage is x mg per every 6 hours for a child with weight 25 lbs i.e. 11.325 kg.

So, $20 \leq \frac{4x}{11.325} \leq 40$

⇒ $20\leq 0.3532x \leq 40$

⇒ $56.625 \leq x \leq 113.25$

So, the right dosage is from 56.625 mg to 113.25 mg of antibiotic to be given every 6 hours for the child with weight 25 lbs. (Answer)

5 0

3 years ago

I need help with this.....

Hatshy [7]

Answer:

55 degrees

Step-by-step explanation:

4 0

3 years ago

A student is asked to factor, if possible, the given expression. The student's response demonstrates which common misconception?

dlinn [17]

Answer:

The student assumed he was asked to factor

(a^2 - b^2)

Which is equal to

= (a+b)*(a-b) (His response)

But, he was asked to simplify

(a^2 + b^2)

Which has imaginary roots, and can be factored as:

(b +ai)(b-ai)

8 0

3 years ago

If the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week

serg [7]

The population after 20 weeks will be 403.42 $x_{0}$ in which $x_{0}$ is the initial population.

Given that the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week.

We are required to find the number of bacteria present after 10 weeks.

let the number of bacteria present at t is x.

So,

dx/dt∝x

dx/dt=kx

1/x dx=k dt

Now integrate both sides.

$\int\limits {1/x} \, dx$ = $\int\limits{k} \, dt$

log x=kt+log c----------1

Put t=0

log $x_{0}$ =0 +log c ( $x_{0}$ shows the population in beginning)

Cancelling log from both sides.

c= $x_{0}$

So put c= $x_{0}$ in 1

log x=kt+log $x_{0}$

log x=log $e^{kt}$ +log $x_{0}$

log x=log $e^{kt}x_{0}$

x= $e^{kt}x_{0}$

We have been given that the population triples in a week so we have to put the value of x=2 $x_{0}$ and t=1 to get the value of k.

2 $x_{0}$ = $e^{k} x_{0}$

2= $e^{k}$

log 2=k

We have to now put the value of t=20 and k=log 2 ,to get the population after 20 weeks.

x= $e^{20log 2}$ $x_{0}$

x= $e^{0.30*2}$ $x_{0}$

x= $e^{6}x_{0}$

x=403.42 $x_{0}$

Hence the population after 20 weeks will be 403.42 $x_{0}$ in which $x_{0}$ is the initial population.

Learn more about growth rate at brainly.com/question/25849702

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The given question is incomplete as the question incudes the following:

Calculate the population after 20 weeks.

5 0

2 years ago